Match each entry in List-I to the correct entry in List-II and choose the correct option.

Evaluating entry (P):

$$\sin^6 x + \cos^4 x = 1, \quad x \in [-\pi, \pi]$$

$$\sin^6 x = 1 - \cos^4 x = (1 - \cos^2 x)(1 + \cos^2 x) = \sin^2 x(1 + \cos^2 x)$$

$$\sin^2 x (\sin^4 x - 1 - \cos^2 x) = 0 \implies \sin^2 x (\sin^4 x - 1 - (1 - \sin^2 x)) = 0$$

$$\sin^2 x (\sin^4 x + \sin^2 x - 2) = 0 \implies \sin^2 x (\sin^2 x + 2)(\sin^2 x - 1) = 0$$

$$\sin x = 0 \quad \text{or} \quad \sin^2 x = 1 \implies \cos x = 0$$

$$x = 0, \pm\pi, \pm\frac{\pi}{2} \implies n(P) = 5 \implies \text{(P)} \rightarrow \text{(5)}$$

Evaluating entry (Q):

$$\sin^2 x + \cos^6 x = 1, \quad x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$

$$\cos^6 x = 1 - \sin^2 x = \cos^2 x \implies \cos^2 x (\cos^4 x - 1) = 0$$

$$\cos x = 0 \quad \text{or} \quad \cos^2 x = 1 \implies \sin x = 0$$

$$x = 0, \pm\frac{\pi}{2} \implies n(Q) = 3 \implies \text{(Q)} \rightarrow \text{(3)}$$

Evaluating entry (R):

$$\cos^2\left(\frac{x}{2}\right) - \sin^2 x = \frac{1}{2}, \quad x \in [-\pi, \pi]$$

$$\frac{1 + \cos x}{2} - (1 - \cos^2 x) = \frac{1}{2} \implies 1 + \cos x - 2 + 2\cos^2 x = 1$$

$$2\cos^2 x + \cos x - 2 = 0 \implies \cos x = \frac{-1 \pm \sqrt{17}}{4}$$

$$\text{Since } \frac{-1 - \sqrt{17}}{4} < -1, \quad \cos x = \frac{\sqrt{17} - 1}{4} \approx 0.78$$

$$x = \pm\cos^{-1}\left(\frac{\sqrt{17}-1}{4}\right) \implies n(R) = 2 \implies \text{(R)} \rightarrow \text{(2)}$$

Evaluating entry (S):

$$6\sin^2\left(\frac{x}{2}\right) - \cos 3x = 3, \quad x \in [-2\pi, 2\pi]$$

$$3(1 - \cos x) - \cos 3x = 3 \implies 3 - 3\cos x - (4\cos^3 x - 3\cos x) = 3$$

$$-4\cos^3 x = 0 \implies \cos x = 0$$

$$x = \pm\frac{\pi}{2}, \pm\frac{3\pi}{2} \implies n(S) = 4 \implies \text{(S)} \rightarrow \text{(4)}$$